What is the solution for y^{''}+4y^'+3y=e^{3x} ?

The solution for y^{''}+4y^'+3y=e^{3x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+1/24 e^{3x}

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y^{''}+4y^'+3y=e^{3x}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{3 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{24} e^{3 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = e^{3 x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{24} e^{3 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{24} e^{3 x}

d, d (p) = 480 - 3 p

y^{^{''}} + 4 y = 3 \cdot sin (2 x)

y^{^{''}} - y = 5 x + 2

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{2 x} - 2 x

\frac{d ^{3} y}{d x ^{3}} + y = cos (x)

What is the solution for y^{''}+4y^'+3y=e^{3x} ?
The solution for y^{''}+4y^'+3y=e^{3x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+1/24 e^{3x}